Chapter 8, Problem 8.69AP

Calculate the value of [OH⁻] from the given [H₃O⁺] and label the solution as acidic or basic.

1. a. 10⁻⁸ M
2. b. 10⁻¹⁰ M
3. c. 3.0 × 10⁻⁴ M
4. d. 2.5 × 10⁻¹¹ M

Interpretation Introduction

Interpretation:

The ${\left[\text{OH}\right]}^{\text{-}}$ value of ${\text{10}}^{\text{-8}}\text{M}$ has to be calculated and the solution has to be determined as acidic or basic.

Concept Introduction:

The ion-product constant for water $\left({\text{K}}_{\text{w}}\right)$ is calculated using the formula,

  ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

The value of ${\text{K}}_{\text{w}}$ is $\text{1}{\text{.0×10}}^{\text{-14}}$ for all aqueous solution at ${\text{25}}^{\text{o}}\text{C}\text{.}$

For a solution to be acidic, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{>}\left[{\text{OH}}^{\text{-}}\right].$

For a solution to be basic, $\left[{\text{OH}}^{\text{-}}\right]\text{>}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right].$

Answer to Problem 8.69AP

The value of $\left[{\text{OH}}^{\text{-}}\right]$ is ${\text{10}}^{\text{-6}}\text{M}\text{.}$

The solution is basic.

Explanation of Solution

Given,

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{=10}}^{\text{-8}}\text{M}\\ {\text{K}}_{\text{w}}\text{=1}{\text{.0×10}}^{\text{-14}}\end{array}$

The concentration of ${\text{OH}}^{\text{-}}$ is calculated as,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{10}}^{\text{-8}}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=1}{\text{.0×10}}^{\text{-6}}\text{M}\end{array}$

The value of $\left[{\text{OH}}^{\text{-}}\right]$ is ${\text{10}}^{\text{-6}}\text{M}\text{.}$

The solution is said to be basic, since the concentration of ${\text{OH}}^{\text{-}}$ is greater than concentration of hydronium ions.

Interpretation Introduction

Interpretation:

The ${\left[\text{OH}\right]}^{\text{-}}$ value of ${\text{10}}^{\text{-10}}\text{M}$ has to be calculated and the solution has to be determined as acidic or basic.

Concept Introduction:

Refer to part (a).

Answer to Problem 8.69AP

The value of $\left[{\text{OH}}^{\text{-}}\right]$ is ${\text{10}}^{\text{-4}}\text{M}\text{.}$

The solution is basic.

Explanation of Solution

Given,

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{=10}}^{\text{-10}}\text{M}\\ {\text{K}}_{\text{w}}\text{=1}{\text{.0×10}}^{\text{-14}}\end{array}$

The concentration of ${\text{OH}}^{\text{-}}$ is calculated as,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{10}}^{\text{-10}}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=1}{\text{.0×10}}^{\text{-4}}\text{M}\end{array}$

The value of $\left[{\text{OH}}^{\text{-}}\right]$ is ${\text{10}}^{\text{-4}}\text{M}\text{.}$

The solution is said to be basic, since the concentration of ${\text{OH}}^{\text{-}}$ is greater than concentration of hydronium ions.

Interpretation Introduction

Interpretation:

The ${\left[\text{OH}\right]}^{\text{-}}$ value of $\text{3}\text{.0}\times {\text{10}}^{\text{-4}}\text{M}$ has to be calculated and the solution has to be determined as acidic or basic.

Concept Introduction:

Refer to part (a).

Answer to Problem 8.69AP

The value of $\left[{\text{OH}}^{\text{-}}\right]$ is $\text{3}\text{.33}\times {\text{10}}^{\text{-11}}\text{M}\text{.}$

The solution is acidic.

Explanation of Solution

Given,

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=3}\text{.0}\times {\text{10}}^{\text{-4}}\text{M}\\ {\text{K}}_{\text{w}}\text{=1}{\text{.0×10}}^{\text{-14}}\end{array}$

The concentration of ${\text{OH}}^{\text{-}}$ is calculated as,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{3.0\times {\text{10}}^{\text{-4}}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=3}{\text{.33×10}}^{\text{-11}}\text{M}\end{array}$

The value of $\left[{\text{OH}}^{\text{-}}\right]$ is $\text{3}\text{.33}\times {\text{10}}^{\text{-11}}\text{M}\text{.}$

The solution is said to be acidic, since the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is greater than the concentration of hydroxide ions.

Interpretation Introduction

Interpretation:

The ${\left[\text{OH}\right]}^{\text{-}}$ value of $\text{2}\text{.5}\times {\text{10}}^{\text{-11}}\text{ M}$ has to be calculated and the solution has to be determined as acidic or basic.

Concept Introduction:

Refer to part (a).

Answer to Problem 8.69AP

The value of $\left[{\text{OH}}^{\text{-}}\right]$ is $\text{4}{\text{.0×10}}^{\text{-4}}\text{M}\text{.}$

The solution is basic.

Explanation of Solution

Given,

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=2}\text{.5}\times {\text{10}}^{\text{-11}}\text{ M}\\ {\text{K}}_{\text{w}}\text{=1}{\text{.0×10}}^{\text{-14}}\end{array}$

The concentration of ${\text{OH}}^{\text{-}}$ is calculated as,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{2.5\times {\text{10}}^{\text{-11}}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=4}{\text{.0×10}}^{\text{-4}}\text{M}\end{array}$

The value of $\left[{\text{OH}}^{\text{-}}\right]$ is $\text{4}{\text{.0×10}}^{\text{-4}}\text{M}\text{.}$

The solution is said to be basic, since the concentration of ${\text{OH}}^{\text{-}}$ is greater than concentration of hydronium ions.